$\overline{AC} = 16$ $\overline{BC} = {?}$ $A$ $C$ $B$ $16$ $?$ $ \sin( \angle BAC ) = \dfrac{15}{17}, \cos( \angle BAC ) = \dfrac{8}{17}, \tan( \angle BAC ) = \dfrac{15}{8}$
Solution: $\overline{BC}$ is the opposite to $\angle BAC$ $\overline{AC}$ is adjacent to $\angle BAC$ SOH CAH TOA We know the adjacent side and need to solve for the opposite side so we can use the tan function (TOA) $ \tan( \angle BAC ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\overline{BC}}{\overline{AC}}= \frac{\overline{BC}}{16} $ Since we have already been given $\tan( \angle BAC )$ , we can set up a proportion to find $\overline{BC}$ $ \tan( \angle BAC ) = \dfrac{15}{8} = \frac{\overline{BC}}{16}$ Simplify. $\overline{BC} = 30$